Review Notes for Linear Algebra True or False Last Updated: February 22, 2010

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1 Review Notes for Linear Algebra True or False Last Updated: February 22, 2010 Chapter 4 [ Vector Spaces 4.1 If {v 1,v 2,,v n } and {w 1,w 2,,w n } are linearly independent, then {v 1 +w 1,v 2 +w 2,,v n +w n } is linearly Let {e 1,e 2 } be the standard basis of R 2. Take {v 1,v 2 } = {e 1,e 2 } and {w 1,w 2 } = { e 1, e 2 }. Then {v 1,v 2 } and {w 1,w 2 } are linearly But then {v 1 + w 1,v 2 + w 2 } = {0,0} is of course linearly dependent. 4.2 If {v 1,v 2,,v n } and {w 1,w 2,,w n } are linearly dependent, then {v 1 +w 1,v 2 +w 2,,v n +w n } is linearly dependent. Take {v 1,v 2 } = {e 1,0} and {w 1,w 2 } = {0,e 2 }. Then {v 1,v 2 } is linearly dependent because 0 e = 0, and {w 1,w 2 } is linearly dependent because e 2 = 0. But then {v 1 + w 1,v 2 + w 2 } = {e 1,e 2 } is of course linearly 4.3 If {v 1,v 2,,v n } is linearly independent, and {v n,v n+1,,v n+k } is linearly independent, then {v 1,v 2,,v n+k } is linearly Take {v 1,v 2 } = {(1, 0),(0, 1)} and {v 2,v 3 } = {(0, 1), (1,1)}. Then v 3 = v 1 + v If {v 1,v 2,,v n+k } is linearly independent, then {v 1,v 2,,v n } is linearly True. If any nonempty subset of S = {v 1,v 2,,v n+k } is linearly dependent, then S must also be linearly dependent. 4.5 If {v 1,v 2,,v n+k } is linearly dependent, then {v 1,v 2,,v n } is linearly dependent. {(0, 1), (1, 0), (1, 1)} are linearly dependent but {(0, 1), (1, 0)} are linearly 4.6 If any two of v 1, v 2, v 3 are linearly independent, then the three vectors v 1, v 2, v 3 are linearly Take v 1 = (1,0), v 2 = (0,1), v 3 = (1, 1). Then three vectors in R 2 must be linearly dependent. 4.7 If the three vectors v 1, v 2, v 3 are linearly independent, then any two of v 1, v 2, v 3 are linearly True. Follows from Let S be a set of k vectors in R n. If k < n, then S is linearly Take v 1 = (1,1, 1), v 2 = (2,2, 2) in R 3. Then S = {v 1,v 2 } is linearly dependent. 4.9 If rank A m n = m, then the columns of A are linearly [ 1 A = with ranka = 2. A has a zero column = not all columns of A are pivot = columns of A are linearly dependent If rank A m n = n, then the columns of A are linearly True. rank A = n = all columns of A are pivot = columns of A are linearly 1

2 4.11 If rank A m n = m, then the rows of A are linearly True. rank A = m = all rows of A are pivot = rows of A are linearly 4.12 If rank A m n = n, then the rows of A are linearly A = with ranka = 2. A has a zero row = not all rows of A are pivot = rows of A are linearly dependent The columns of a matrix A are linearly independent if the equation Ax = 0 has the trivial solution. if and only if the equation Ax = 0 has only the trivial solution. The columns of a matrix A are linearly independent 4.14 The columns of any 4 5 matrix are linearly If a set contains more vectors than there are entries in each vector, then the set is linearly dependent. That is, any set {v 1,v 2,,v m} in R n is linearly dependent if m > n If rank A m n = m, then the columns of A span R m. True. rank A = m = all rows of A are pivot = Ax = b has solutions for all b R m = b is a linear combination of columns of A for all b = columns of A span R m If rank A m n = n, then the columns of A span R m. A = with ranka = 2. We can find b = R 3 such that 1 b is not a linear combination of 1 0, 0 1 = columns of A do not span R If b is in the span of {v 1,v 2,,v n+1 }, then b is in the span of {v 1,v 2,,v n }. Let {e 1,e 2 } be the standard basis of R 2. Then (1,1) span {e 1,e 2 } but (1,1) span {e 1 } If b is in the span of {v 1,v 2,,v n }, then b is in the span of {v 1,v 2,,v n+1 }. True. b = a 1 v a nv n for some a i. Then b = a 1 v a nv n + 0v n If v 1, v 2,, v n are in R m, then span {v 1,v 2,,v n } is the same as the column space of the matrix [ v1 v 2 v n. True. If A = [ v 1 v 2 v n, with the columns in R m, then ColA is the same as span {v 1,v 2,,v n}. The column space of an m n matrix is a subspace of R m The row space of A is the same as the column space of A t. True. If A is an m n matrix, each row of A has n entries and thus can be identified with a vector in R n. The set of all linear combinations of the row vectors is called the row space of A and is denoted by RowA. Each row has n entries, so RowA is a subspace of R n. Since the rows of A can be identified with the columns of A t, we could also write ColA t in place of RowA. 2

3 4.21 If v 1, v 2,, v n span R n, then {v 1,v 2,,v n } is linearly True. Take A = [ v 1 v 2 v n. Then A is square. Now, v1, v 2,, v n span R n = all rows of A are pivot = ranka = n = all columns of A are pivot = {v 1,v 2,,v n} is linearly 4.22 If v 1, v 2,, v n span R n, then {v 1,v 2,,v n } is a basis of R n. True. Follows from If columns of A 5 3 are linearly independent, then rows of A span R 3. True. Columns of A are linearly independent = all columns of A are pivot = all rows of A t are pivot = ranka t = 3. Then for any b R 3, rank [ A t b 3 6 = rankat = 3 = b is a linear combination of columns of A t for any b R 3 = b is a linear combination of rows of A for any b R 3 = rows of A span R If v 1, v 2,, v n span V, then dim V n. True. If {v 1,v 2,,v n} is linearly independent, then {v 1,v 2,,v n} is a basis of the span V. Thus, dim V = number of vectors in a basis of V = n. If {v 1,v 2,,v n} is linearly dependent, then dim V < n If v 1, v 2,, v n span V, then v 1, v 2,, v n, v n+1,, v n+k span V. True. For b V, there exist scalars c 1, c 2,, c n such that b = c 1 v 1 + c 2 v c nv n = c 1 v 1 + c 2 v c nv n + 0v n v n+k = b can be expressed as a linear combination of v 1, v 2,, v n, v n+1,, v n+k. In other words, v 1, v 2,, v n, v n+1,, v n+k span V If v 1, v 2,, v n do not span V, then v 1, v 2,, v n, v n+1 do not span V. v 1 = (1, 0,0), v 2 = (0, 1,0) do not span R 3, but v 1 = (1,0, 0), v 2 = (0,1, 0), v 3 = (0,0, 1) span R If v 1, v 2,, v n span V, then {v 1,v 2,,v n } is a basis of V. The vectors can be linearly dependent. Let {e 1, e 2, e 3 } be the standard basis of R 3. Then {e 1, e 2, e 1 + e 2 } span a plane in R 3 but {e 1, e 2, e 1 + e 2 } is not a basis (vectors being linearly dependent) If {v 1,v 2,,v n } is a basis of R n and A n n is invertible, then {Av 1,Av 2,,Av n } is also a basis of R n. True. Let B n n = [ [ [ Av 1 Av n = A v1 v n = AT, where T = v1 v n. By given, n n {v 1,,v n} is a basis of R n = all rows as well as all columns of T are pivot = T is row equivalent to I n = T is invertible. Hence, B is also invertible. It follows that, for Av 1,,Av n R n, {Av 1,Av 2,,Av n} is a basis of R n The functions 1, sint, cos t are linearly True. Suppose c 1 + c 2 sin t + c 3 cos t = 0. Then t = 0 = c 1 + c 3 = 0; t = π/2 = c 1 + c 2 = 0; t = π = c 1 c 3 = 0. Hence, c 1 = c 2 = c 3 = 0. By definition, the functions are linearly 4.30 The functions 1, sin 2 t, cos 2 t are linearly Linear dependence relation is: 1 sin 2 t cos 2 t = 0. 3

4 4.31 A homogeneous equation is always consistent. True. Since A0 = 0, the equation Ax = 0 always has the zero solution x = If x is a nontrivial solution of Ax = 0, then every entry in x is nonzero. [ 1 x = 0 has a nontrivial solution x = The set of all solutions of a system of m homogeneous equations in n unknowns is a subspace of R m. The null space of an m n matrix A is a subspace of R n. Equivalently, the set of all solutions of a system Ax = 0 of m homogeneous linear equations in n unknowns is a subspace of R n The columns of an invertible n n matrix from a basis for R n. True. The columns of an invertible n n matrix from a basis for R n because they are linearly independent ( all columns of A are pivot) and span R n ( all rows of A are pivot) If ranka 7 5 = 5, then the dimension of the null space of A is 0. True. ranka = 5 = all columns of A are pivot = no nonpivot column of A = no free variable = Ax = 0 has only trivial solution = NulA def. = {x : Ax = 0} = {0} = NulA has no basis = dim (NulA) = The dimension of NulA is the number of variables in the equation Ax = b. By definition, the dimension of NulA is the number of vectors in a basis of NulA. The number of vectors in a basis of NulA is equal to the number of nonpivot columns of A which must be strictly smaller than the number of columns of A (in other words, which must be strictly smaller than the number of variables) The dimension of ColA is the number of pivot columns of A. the dimension of ColA is just the number of pivot columns of A, that is the rank of A. True. Since the pivot columns of A form a basis for ColA, 4.38 The dimensions of ColA and NulA add up to the number of columns of A. True. The nonpivot columns of A correspond to the free variables in Ax = 0. Thus the dimension of Nul A is the number of nonpivot columns of A. Since the number of pivot columns plus the number of nonpivot columns of A is exactly the number of columns, the dimensions of Col A and Nul A have the useful connection: If a matrix A has n columns, then rank A + dim Nul A = n If Ax = 0 has only trivial solution, then the columns of A form a basis of ColA. True. Ax = 0 has only trivial solution = no free variable = no nonpivot column of A = all columns of A are pivot = columns of A are linearly independent = columns of A form a basis of ColA If Ax = 0 has only trivial solution, then the rank of A is the number of columns of A. True. Ax = 0 has only trivial solution = all columns of A are pivot = ranka = number of columns of A. 4

5 4.41 If Ax = 0 has only trivial solution, then the rank of A is the number of rows of A. A = Then Ax = 0 has only trivial solution but ranka = If the columns of an m n matrix A span R m, then the equation Ax = b is consistent for each b in R m. True. Columns of A span R m = b is a linear combination of columns of A for all b R m = Ax = b has solutions for all b R m If A is an m n matrix whose columns do not span R m, then the equation Ax = b is inconsistent for some b in R m. True. Columns of A do not span R m = Ax = b has no solution for some b R m. = b cannot not be a linear combination of columns of A for some b R m 4.44 If A is an m n matrix and if the equation Ax = b is inconsistent for some b in R m, then A cannot have a pivot position in every row. = not all rows of A are pivot. True. Ax = b has no solution for some b in R m = b is pivot in [ A b for some b in R m 4.45 If Ax = b has solutions for all b, then the columns of A form a basis of ColA. [ [ b1 A = and b =. Then Ax = b has solutions for all b but columns of A are linearly dependent b 2 (column 3 = column 1 + column 2). Thus, columns of A do not form a basis of ColA If B is a row echelon form of a matrix A, then the pivot columns of B form a basis for ColA. A = = B. 1 0 However, the pivot columns (the first and third) of B do not span ColA because, for example, (0, 0,1) cannot be a linear combination of the pivot columns of B If Ax = b has solutions for all b, then the rank of A is the number of columns of A. Ax = b has solutions for all [ b = all rows of A are pivot = ranka = number of rows of A 1 ranka = number of columns of A. Take A = If Ax = b has solutions for all b, then the rank of A is the number of rows of A. True. Ax = b has solutions for all b = all rows of A are pivot = ranka = number of rows of A If [ A b is an invertible 4 4 matrix, then b is in the column space of A. [ [ A b is invertible = A b is row equivalent to I = b is pivot = Ax = b has no solution = b is not a linear combination of columns of A = b ColA If [ A b is an 3 5 matrix with ranka = 3, then b is in the column space of A. True. rank A = 3 = all rows of A are pivot = Ax = b has solutions for all b = b Col A. 5

6 4.51 If [ A b is an 4 4 matrix with rank [ A b = ranka, then b is in the column space of A. True. rank [ A b = ranka = b is nonpivot = Ax = b has solutions = b ColA If the vectors v 1, v 2, v 3, v 4 in R 4 are linearly dependent, then the 4 4 matrix A = [ v 1 v 2 v 3 v 4 is not invertible. True. Columns of A are linearly dependent = not all columns of A are pivot = ranka < 4 = row echelon form of A has a zero row = deta = 0 (by 2.31) = A is not invertible If the 4 4 matrix A = [ v 1 v 2 v 3 v 4 is not invertible, then the vectors v1, v 2, v 3, v 4 in R 4 are linearly dependent. True. A is not invertible = A is not row equivalent to I = row echelon form of A has a zero row = ranka < 4 = not all columns of A are pivot = columns of A are linearly dependent If A is an 3 3 matrix with ranka = 3, then ColA = R 3. b ColA for all b R 3 = ColA R 3 = ColA = R 3. True. ranka = 3 = all rows of A are pivot = columns of A span R 3 = 4.55 If A is an 3 3 matrix with ranka = 3, then RowA = R 3. True. ranka = 3 = ranka t = 3 = ColA t = R 3 (by 7.54) = RowA = R If v 1 and v 2 are in R 4 and v 1 is not a scalar multiple of v 2, then {v 1,v 2 } is linearly Choose v 1 0 and v 2 = If v 1, v 2, v 3, v 4 are vectors in R 4 and {v 1,v 2,v 3 } is linearly independent, then {v 1,v 2,v 3,v 4 } is linearly Choose v 4 = v If v 1, v 2, v 3, v 4 are linearly independent vectors in R 4, then {v 1,v 2,v 3 } is linearly True. Refer to If V has a basis of n vectors, then every basis of V must consist of exactly n vectors. True. Let B = {b 1,b 2,,b n} be a basis of V. Suppose that B 1 = {d 1,d 2,,d m} is another basis of V. Recall that a basis is a maximal independent set as well as a minimal spanning set. Now, since B 1 is linearly independent, we have m n. On the other hand, B 1 is a basis and B is linearly independent, we have n m as well. Thus, m = n The span of any vector in R 3 is a line. v = 0 = span {v} = {0} contains origin only The span of any two nonzero vectors in R 3 is a plane. Nonzero u, v in R 3 such that u v = span {u,v} = span {u} = a line. 6

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